Why does an open switch stop a circuit?

[Gear300]

2 capacitors are connected in parallel to a battery and a switch is connected next to the negative terminal of the battery. If the switch is left open, why does the circuit stop functioning. There is no switch next to the positive terminal, and since the electric field goes from positive to negative, shouldn't the circuit technically still function in areas behind the switch.

 

[Mindscrape]

I don't know if I fully understand what you are asking. In a sense the circuit does still function, it depends on what you want to have your circuit do. Perhaps you are using the charge from the capacitors to run a photomultiplier for an IR viewer or something, you only need the battery to be closed for as long as the charging takes (eventually the parasitic resistances will make your viewer dim so you have to charge again).

In another way, the capacitors are now completely isolated from the battery because the charge has nowhere to flow (except the possible load that I mentioned above). Maybe you are worried about it being at the negative terminal, and the fact that we look at the flow of positive current. Think about the actual electrons, which would be flowing in the opposite direction.

 

[Feldoh]

It requires both ends of the battery to charge a capacitor. If you unhook the battery you'll still have the charged capacitors, but you won't be able to recharge them. You could essentially look at it by saying there's still a circuit, however someone took your battery away and so it is no longer part of the circuit.

 

[Mike Cookson]

I realize there are other explanations but here's my contribution...

Current is a the movement of electrons in a conductor. Now like a plumbing system you need somewhere for the water, or electrons, to come from and go to. When you open the switch there is no where for electrons to go. With no batter connected you have no voltage across the capacitors so they won't charge or draw current.

 

[Gear300]

I sort of get what all of you are saying. Its sort of like saying that current stops at the open switch, and so...it stops at all previous points.

 

[Mike Cookson]

In a way yes. Current must have a path to flow, all current that exits the battery has to enter it (for this case anyway). Having the switch is like having a valve in a plumbing system that pumps water round a series of pipes with no output or leaks, when the valve stops the water, i.e. the switch is open, the pump cannot move the water with enough force so there is no flow, which is the current.

Of course there are cases where the 'force' could over come the valve - the electrical alternative would be where the voltage across the switch is high enough to overcome the resistance of the air. (I only included this to show that my plumbing model equivalent makes sense with the pump having enough 'force') I hope this hasn't complicated matters. It's much easier to explain in person.

 

[Gear300]

I get what you're saying. Its sort of how with a high enough electric field, it could produce a discharge.

 

[Mike Cookson]

Yes, If you look at the specification of a switch there will be things like it's useable voltage range etc. (and possibly something akin to a breakdown voltage). Somewhere above the usable range, which might be called the breakdown voltage, the insulation of the air between the two contacts of the switch is not sufficient to stop current and it sparks.

I don't recommend you do this on purpose but sometimes if a light switch is of poor quality when it is pressed the gap between conductors does not open/close as quickly as it should so for a short period the gap between conductors is short enough for the 240V mains (UK) to create a spark.

I have managed to stray off topic, are you satisfied with the answer to your original question?

 

[Gear300]

Yes...you've answered clearly, thanks.